In a book I am studying, while solving something, it does the following thing:

From here $$ e^{-iωτ}=\frac{1}{2}-iω$$

it goes here

$$ τ=\frac{-arg(1/2-iω)}{ω} $$

However I don't get how this happened. I looked up properties to do it my self, but I can't find it. Can anybody tell me what properties where used to do this?

P.S. I hope this question is valid for this Q&A. I have spent 2 hours trying to solve this, something that I think is really easy.

  • 4
    $\begingroup$ We have $e^{i(-\omega \tau)} = \frac 1 2 - i\omega$. By definition of $\mathrm{arg}$, we have $-\omega \tau = \mathrm{arg}(1/2 - i\omega)$. Then you just solve for $\tau$. $\endgroup$ – Trevor Norton Mar 26 '18 at 19:30
  • $\begingroup$ @TrevorNorton now I see what you mean. I was obvious but I still couldn't see it. Thanks $\endgroup$ – Dimitris Pantelis Mar 27 '18 at 17:00

The answer was from @TrevorNorton comment. I am just writing it down for completion reasons.

The property used is ( based on the definitions of $arg()$ and $e^{θi}$ ): $$ arg(e^{θi})=θ $$

So by applying $arg()$ on both sides, we can get the exponential out.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.